Question #249006

Find the mass of ammonia in the 100 cubic feet tank having a pressure of 70 psi at 120 °F.


1
Expert's answer
2021-10-10T15:59:15-0400

Ideal gas law

PV=nRTV=100  ft3=2831.68  LP=70  psi=4.7632  atmT=120  °F=322.039PV=nRT \\ V=100 \;ft^3 = 2831.68 \;L\\ P=70 \;psi = 4.7632 \;atm\\ T= 120 \;°F = 322.039

R=0.082058 L×atm/(K×mol)

n=PVRT=4.7632×2831.680.082058×322.039=510.40  moln = \frac{PV}{RT} \\ = \frac{4.7632 \times 2831.68}{0.082058 \times 322.039} \\ = 510.40 \;mol

M(ammonia) = 17.031 g/mol

m(ammonia) =510.40×17.031=8692.62  g=8.692  kg= 510.40 \times 17.031 = 8692.62 \;g = 8.692 \;kg

Answer: 8.7 kg


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